I'm trying to write precise coordinate and coordinate-free definitions for the covariant and directional covariant derivative of a tensor. I have some working definitions from lecture, but I feel like they could be made better (also I'm not too sure how to write both coordinate and coordinate free definitions for either of these).

Covariant Derivative

The covariant derivative of a tensor field $T$ with respect to a linear connection $\nabla$, is $$\nabla T = \nabla_{E_i}T\otimes \theta^i$$ where $\{E_i\}$ forms a local frame, $\{\theta^i\}$ its dual co-frame, and $\nabla_{E_i}(T)$ is the directional covariant derivative.

Directional Covariant Derivative

Let $T$ and $S$ be tensors, then the directional covariant derivative, $\nabla_X$, has the defining property $$\nabla_X(T\otimes S)=\nabla_XT \otimes S + T\otimes \nabla_XS.$$


1 Answer 1


The wording "directional covariant derivative" is not widely used in the literature, but some authors (e.g. Amari, Information Geometry and Its Applications, p. 117) use it, perhaps, to distinguish from the "total covariant derivative" (see e.g. J.M.Lee, Riemannian Manifolds: An Introduction to Curvature, p. 54), which is a tensor $\nabla T$ of a higher rank, given by $$ \nabla T (\omega_1, \dots, \omega_p, V_1, \dots, V_q, X) = \nabla_X T (\omega_1, \dots, \omega_p, V_1, \dots, V_q) $$ As I get it, if $\nabla T$ denotes the total covariant derivative as above, then $\nabla_{X} T$ is the directional covariant derivative of $T$ in the direction of vector field $X$, and $\nabla_X T (\dots) = \nabla T (\dots, X) $.


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